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Size:  44.7 KB For aii) I don't get why 'At 60ºC enzyme reacts rapidly' would worth a mark. The other two marks is by saying not all substates have reacted and enzymes have denature but I don't get the first marking point.

    My second question is for b), the answer is that all the substrates have been converted into products but how would you know that. Since the substrate conc and enzyme conc are the same. So surely the active site could be the limiting factor and this can cause the curve to flatten but this is not the answer...


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    Hi, let me try to help you - I don't often have time to sleep, but tonight my eyes are drooping slightly so I might not explain v well!

    a) ii) You are are absolutely correct in doubting the validity of this answer - to be perfectly honest, it does not make sense, and is an example of the poor knowledge of some exam paper producers, I regret to say, which is extremely unfair to very capable students - I am sorry, but fact is fact, even if it makes me unpopular!: is this a Q from Edexcel?

    b) When you plot rate of reaction (of which mass of product formed is one indication) on the y axis against time on the x axis, it is a known fact that at normal body temperature (37 degrees C), the graph will level out when the substrate is used up; POINT ONE: the question says that there are FIXED concentrations of substrate and enzyme (meaning the same at each temperature to allow comparison) not the SAME concentration of substrate and enzyme at any one temperature. POINT TWO: the definition of an enzyme specifies that an enzyme speeds up a reaction WITHOUT ITSELF BEING USED UP, so the enzyme cannot be a limiting factor, except that if there was too little enzyme present, the rate of collisions between substrate and enzyme would be reduced overall [leading to reduced binding with active sites], so the graph would be flatter [smaller gradient] and would therefore reach the saturation point later, but at the same height on the y axis, and would still be determined by the complete utilization of the substrate, only this process would take longer.

    So in this case, (surpriingly? ) Edexcel have got it right.

    Hope this makes just a little sense!

    Mukesh (Specialist Biology tutor)
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    (Original post by macpatelgh)
    Hi, let me try to help you - I don't often have time to sleep, but tonight my eyes are drooping slightly so I might not explain v well!

    a) ii) You are are absolutely correct in doubting the validity of this answer - to be perfectly honest, it does not make sense, and is an example of the poor knowledge of some exam paper producers, I regret to say, which is extremely unfair to very capable students - I am sorry, but fact is fact, even if it makes me unpopular!: is this a Q from Edexcel?

    b) When you plot rate of reaction (of which mass of product formed is one indication) on the y axis against time on the x axis, it is a known fact that at normal body temperature (37 degrees C), the graph will level out when the substrate is used up; POINT ONE: the question says that there are FIXED concentrations of substrate and enzyme (meaning the same at each temperature to allow comparison) not the SAME concentration of substrate and enzyme at any one temperature. POINT TWO: the definition of an enzyme specifies that an enzyme speeds up a reaction WITHOUT ITSELF BEING USED UP, so the enzyme cannot be a limiting factor, except that if there was too little enzyme present, the rate of collisions between substrate and enzyme would be reduced overall [leading to reduced binding with active sites], so the graph would be flatter [smaller gradient] and would therefore reach the saturation point later, but at the same height on the y axis, and would still be determined by the complete utilization of the substrate, only this process would take longer.

    So in this case, (surpriingly? ) Edexcel have got it right.

    Hope this makes just a little sense!

    Mukesh (Specialist Biology tutor)
    Hi, thanks a lot for your help but I don't get what you mean about the bits in bold. Doesn't the first part in bold basically mean the same thing? Same concentration temperature at each temperature and same concentration at any temperature, what is the difference?


    'except that if there was too little enzyme present' -> This is what I mean by the enzyme concentration is a limiting factor because the active sites are occupied.

    Any clarification will be great. Thanks
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    Hi, let me explain both points in a different way:

    TEMPERATURE POINT: When they say fixed concentrations of enzyme and of substrate, I would interpret it as the same concentration of substrate at, (let us say, off the top of my head, 0.05M [molar concentration) i.e. 0.5M at 60C, 0.5M at 37C AND 0.5M at 25C (the same conc-n of substrate at the 3 temp-s in degrees centigrade).AND (let us say, as example again), 1000 IU of enzyme at 60C, 1000 IU at 37C AND 1000 IU at 25C.

    From the way you have put in the explanation of your Q, it seems like you are interpreting it as the same concentration of enzyme AS the concentration of substrate i.e. conc-n of substate = conc-n of enzyme. I don't think this is likely since enzymes have their catalytic effect on biological reactions at very low concentrations, therefore being quantified (usually) in IU (International Units) rather than molar conc-n [which is the case for substrates, or other reactants AND for products].

    The effect of this distinction between the two interpretations (which also answers your second point [in bold]) is that although the enzyme concentration [being finite, and not unlimited], will determine the number of active sites at any particular moment in time [therefore controlling the RATE of reaction, as you correctly state], this is a dynamic process whereby once a molecule of enzyme has had a molecule of substrate bound to its active site AND THIS HAS BEEN CONVERTED TO A MOLECULE(S) of PRODUCT(S), the active site BECOMES FREE to bind to another molecule of substrate. THE term limiting factor is confusing, but as far as my understanding goes, a limiting factor normallly refers to (at least in the context of this Q) the factor that makes the reaction reach a plateau, and go NO FURTHER i.e. NO MORE PRODUCT formed rather than a factor that limits only the speed of the reaction.

    I hope this makes it easier to understand that when the substrate is ALL USED UP no more product can be formed so the graph line BECOMES HORIZOTAL, (AND THAT IS WHAT THE Q ASKS YOU TO EXPLAIN [explain Q of course means "WHY?"]); The enzyme is NEVER (within reason) all used up, so it can only make the graph less steep going rightwards, but NOT horizontal as is depicted in the diagram for 37C. The ONLY exception to this (sufficient for A level - KEEP IT SIMPLE!) is the example of the 60C graph, where the enzyme is "USED UP" because it is destroyed (denatured), and so the graph does level out even though there is substrate remaining (we know this because the graph of 37C flattens out at a higher product conc-n, which tells us that there was more substrate available at 60C, but no enzyme to convert it to product.

    Sorry for this "PhD thesis" but I felt it necesssary to go through ALL aspects.

    Regards, Mukesh.
 
 
 
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